In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralise 10cm3 of 1.25 molar sulphuric acid?
- 5cm3
- 10cm3
- 20cm3
- 25cm3
-
50cm3
The correct answer is: E
Explanation
\(2NaOH + H_{2}SO_{4} = Na_{2}SO_{4} + 2H_{2}O\)
V x 0.5M = 2(10 x 1.25)
0.5V = 2 x 12.5
0.5V = 25
V = \(\frac{25}{0.5}\) = \(50cm^{3}\)